Corollary 2

Suppose that the diagram (4) is commutative and has exact rows. Under these conditions:

(i) If $g$ is injective and if $f$ and $v$ are surjective, then $h$ is injective. (ii) If $g$ is surjective and if $h$ and $u'$ are injective, then $f$ is surjective.

To prove (i), consider the diagram

where $f'$ is the mapping having the same graph as the restriction of $g$ to $u(\mathbf{M})$, $w$ and $w'$ being the canonical injections; it is clear that this diagram is commutative and has exact rows. Moreover $w'$ is injective, and by assumption $v$ is surjective; we therefore have, by prop. 2 (iii), an exact sequence

since $g$ is injective and $f'$ is surjective, we therefore have $\mathrm{Ker}\,(h) = 0$.

To prove (ii), consider the diagram

where this time $h'$ is the mapping having the same graph as the restriction of $h$ to $\nu(N)$, and $w$ and $w'$ have respectively the same graphs as $\nu$ and $\nu'$; this diagram is commutative and its rows are exact. Moreover $w$ is surjective and by assumption $u'$ is injective; we therefore have, by prop. 2 (iii), an exact sequence

since $g$ is surjective and $h'$ is injective, we therefore have $\operatorname{Coker}(f)=0$.

Corollary 3 (Five Lemma). — Consider a commutative diagram of $A$-modules

where the rows are exact.

(i) If $f_2$ and $f_4$ are injective and $f_1$ surjective, $f_3$ is injective.

(ii) If $f_2$ and $f_4$ are surjective and $f_5$ injective, $f_3$ is surjective.

In particular, if $f_1$, $f_2$, $f_4$ and $f_5$ are isomorphisms, the same is true of $f_3$.

To prove (i), put $\widetilde M_2=\operatorname{Coker}(u_1)$, $\widetilde M'_2=\operatorname{Coker}(u'_1)$ and denote by $\widetilde f_2:\widetilde M_2\to\widetilde M'_2$ the mapping deduced from $f_2$. It follows from cor. 2 (i) that $\widetilde f_2$ is injective. Applying cor. 1 (i) to the diagram

where $\widetilde u_2$ and $\widetilde u'_2$ are deduced from $u_2$ and $u'_2$, we see that $f_3$ is injective.

To prove (ii), set $\widetilde M_4=\operatorname{Ker}(u_4)$, $\widetilde M'_4=\operatorname{Ker}(u'_4)$ and denote by $\widetilde f_4:\widetilde M_4\to\widetilde M'_4$ the mapping induced by $f_4$. It follows from cor. 2 (ii) that $\widetilde f_4$ is surjective. Applying cor. 1 (ii) to the diagram

where $\widetilde u_3$ and $\widetilde u'_3$ have the same graph as $u_3$ and $u'_3$, one sees that $f_3$ is surjective.